Problem 29 - A Graphical Approach to College Algebra

Question

For each pair of functions, (a) find \((f+g)(x),(f-g)(x),\) and \((f g)(x)\) (b) give the domains of the functions in part (a); (c) find \(\frac{f}{8}\) and give its domain; (d) find \(f \circ g\) and give its domain; and (e) find \(g \circ f\) and give its domain. Do not use a calculator. $$f(x)=\sqrt{x^{2}+3}, g(x)=x+1$$

Step-by-Step Solution

Answer

Functions are defined for all real \(x\); except for division where \(x \neq -1\).

Step 1: Find the Sum of Functions

To find the sum \((f+g)(x)\), we add the two functions: \[(f+g)(x) = f(x) + g(x) = \sqrt{x^2+3} + (x+1)\]

Step 2: Calculate the Difference of Functions

To find the difference \((f-g)(x)\), we subtract \(g(x)\) from \(f(x)\): \[(f-g)(x) = f(x) - g(x) = \sqrt{x^2+3} - (x+1)\]

Step 3: Compute the Product of Functions

The product \((f\cdot g)(x)\) is found by multiplying \(f(x)\) and \(g(x)\): \[(f \cdot g)(x) = f(x) \cdot g(x) = \sqrt{x^2+3} \cdot (x+1)\]

Step 4: Determine Domains of (f+g)(x), (f-g)(x), (f·g)(x)

The domain for these functions is where both \(f(x)\) and \(g(x)\) are defined. Here, \(f(x)\) is defined for all \(x\) because \(x^2 + 3\) is always non-negative. \(g(x)\) is defined for all real \(x\). Thus, the domain is all real numbers \(x \in \mathbb{R}\).

Step 5: Find the Quotient \(\frac{f}{g}\) and Its Domain

Calculate the quotient \(\frac{f}{g}\): \[\frac{f}{g}(x) = \frac{\sqrt{x^2+3}}{x+1}\].The domain excludes any \(x\) that makes the denominator zero, so \(x eq -1\). The domain is \(x \in \mathbb{R} \setminus \{-1\}\).

Step 6: Determine \(f \circ g\) and Its Domain

To find \(f \circ g\), use \(f(g(x))\): \[f(g(x)) = f(x+1) = \sqrt{(x+1)^2 + 3}\].Since \(f(x)\) is defined for all real \(x\), \(f \circ g\) is defined for \(x \in \mathbb{R}\).

Step 7: Determine \(g \circ f\) and Its Domain

To find \(g \circ f\), use \(g(f(x))\): \[g(f(x)) = g(\sqrt{x^2+3}) = \sqrt{x^2+3} + 1\].The function \(g \circ f\) is defined for all real \(x\). Therefore, the domain is \(x \in \mathbb{R}\).

Key Concepts

Domain of a Function

The domain of a function is all the possible input values (usually represented as \( x \)) for which the function is defined. Simply speaking, it's the set of all \( x \) values that you can put into a function without running into any problems, like dividing by zero or taking the square root of a negative number. In this context:

• For \( f(x) = \sqrt{x^2+3} \), the expression inside the square root \( x^2 + 3 \) is always non-negative for every real number since \( x^2 \geq 0 \). Thus, \( f(x) \) is defined for all real numbers \( x \).
• Similarly, \( g(x) = x+1 \) is a linear function and is defined for all real numbers\( x \).

As both functions are defined for all real numbers, the domain of the functions \( (f+g)(x) \), \( (f-g)(x) \), and \((f \cdot g)(x)\) also includes all real numbers.

Function Composition

Function composition involves applying one function to the results of another. This is typically denoted as \( f \circ g \), which means we first apply \( g \) then apply \( f \) to the result. Here's a breakdown:

• For \( f \circ g \), you replace the \( x \) in \( f(x) \) with \( g(x) \), i.e., compute \( f(g(x)) = \sqrt{(x+1)^2 + 3} \). The domain here depends on where \( g(x) \) fits \( f \). Since both are valid for all \( x \), so is \( f \circ g \).
• For \( g \circ f \), you substitute \( f(x) \) into \( g(x) \), i.e., \( g(f(x)) = \sqrt{x^2+3} + 1 \). This operation also results in a valid function for all real number inputs, aligning the domain with all \( x \).

When composing functions, always check that the output of the first function agrees with the input requirements of the second. However, in this case, both compositions are indeed defined for all real numbers.

Algebraic Operations on Functions

Algebraic operations on functions involve combining functions using operations such as addition, subtraction, multiplication, and division. Let’s delve into these operations:

• Addition (\(f+g\)): The combined function \((f+g)(x) = \sqrt{x^2+3} + (x+1)\) adds two functions. The domain remains all real numbers as explained before, because both \( f(x) \) and \( g(x) \) are defined everywhere.
• Subtraction (\(f-g\)): Having \((f-g)(x) = \sqrt{x^2+3} - (x+1)\), the domain does not change from the original functions, staying within all real numbers.
• Multiplication (\(f \cdot g\)): The function multiplication \((f \cdot g)(x) = \sqrt{x^2+3} \cdot (x+1)\) still holds true across all real numbers since neither function restricts any inputs.
• Division (\( \frac{f}{g} \)): This operation, \( \frac{f}{g}(x) = \frac{\sqrt{x^2+3}}{x+1}\), introduces a restriction. Here, \( x \) cannot be \(-1\) because it makes the denominator zero, potentially causing undefined behavior. So the domain becomes all real numbers except \( x = -1 \).

When working with algebraic operations on functions, always keep an eye on conditions, like division by zero, that might restrict your domain.